Let $\frac{1}{2}+\frac{1}{3}+...+\frac{1}{121}=\frac{p}{q}$ where p,q are coprime integer couple. Prove $p\equiv 50 \pmod {121}.$
What im guessing about is that wolstenholme will do its job here but i don't exactly know how i apply it
Please give motivation of proof if possible if any...
Note that the sum of the left hand side, when combined into one fraction, has a numerator that consists of products of denominators (all but one). That means that each of the terms of the sum in the numerator has a factor of 121, except for one: $120!$. So, All you need to show is that $120!\equiv50\pmod{121}$.